# Possible remainders when $3^{4n-2}+ 2^{6n-3} + 1$ is divied by $17$

The numbers in the form of $3^{4n-2}+ 2^{6n-3} + 1$, where $n$ is a positive integer, when divided by $17$, has possible remainders?

Note that $$3^{4n-2} + 2^{6n-3} + 1 = 9^{2n-1} + 8^{2n-1} + 1$$ Now $(9+8) \vert \left(9^{2n-1} + 8^{2n-1} \right), \,\,\, \forall n \in \mathbb{Z}^+$ (Why?). Hence, the remainder is $1$.
• (WHY)'s answer: If $m$ is an odd positive integer (as $m=2n-1$), then $X^m+Y^m = (X+Y)(X^{m-1}-X^{m-2}Y+X^{m-3}Y^2\mp\cdots+X^2Y^{m-3}-XY^{m-2}+Y^{m-1})$. Here $X=9$, $Y=8$. Thanks to ccorn – Silent Apr 6 '14 at 13:20
$9^{2n-1}+8^{2n-1}+1 \equiv x (\mod 17)$
$9^{2n-1}\equiv (17-8)^{2n-1}(\mod 17) \implies 9^{2n-1}\equiv 17k-8^{2n-1}(\mod 17)$
$(a-b)^k=a^k$-$k \choose 2$ $a^{k-1} \cdot b+\dots (-1)^k(b^k)$, here $k \in odd$
$8^{2n-1}\equiv 8^{2n-1} (\mod 17) \implies 9^{2n-1}+8^{2n-1} \equiv0 (\mod 17)$